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SOLUTION 7: We are given the function f(x)=2x+√x−1 and the interval [1,5]. This function is continuous on the closed interval [1,5] since it is the sum of continuous functions y=2x (polynomial) and y=√x−1 (functional composition of continuous functions y=x−1 (polynomial) and y=√x (well-known). ). The derivative of f is
f′(x)=2+(1/2)(x−1)−1/2=2+12√x−1
We can now see that f is differentiable on the open interval (1,5). The assumptions of the Mean Value Theorem have now been met. Let's apply the Mean Value Theorem and find all possible values of c in the open interval (1,5). Then
f′(c)=f(5)−f(1)5−1 ⟶
2+12√c−1=(2(5)+√5−1 )−(2(1)+√1−1 )4 ⟶
2+12√c−1=(10+2)−(2+0)4 ⟶
2+12√c−1=52 ⟶
12√c−1=12 ⟶
√c−1=1 ⟶
(√c−1 )2=12 ⟶
c−1=1 ⟶
c=2
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